# Question 7, Exercise 1.1

Solutions of Question 7 of Exercise 1.1 of Unit 01: Complex Numbers. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

If ${{z}_{1}}=1+2i$ and ${{z}_{2}}=2+3i$, evaluate $|{{z}_{1}}+{{z}_{2}}|$.

We know that $z_1=1+2i$ and $z_2=2+3i$, then \begin{align} {{z}_{1}}+{{z}_{2}}&=1+2i+2+3i\\ &=1+2+2i+3i\\ &=3+5i \end{align} Now \begin{align} |z_1+z_2|&=\sqrt{3^2+5^2}\\ &=\sqrt{9+25}\\ &=\sqrt{34}\end{align}

If ${{z}_{1}}=1+2i$ and ${{z}_{2}}=2+3i$, evaluate $|{{z}_{1}}{{z}_{2}}|$.

We know that $z_1=1+2i$ and $z_2=2+3i$, then \begin{align} {{z}_{1}}{{z}_{2}}&=\left( 1+2i \right)\times \left( 2+3i \right)\\ &=\left( 2-6 \right)+\left( 3+4 \right)i\\ &=-4+7i. \end{align} Now \begin{align} |z_1 z_2|&=\sqrt{(-4)^2+7^2}\\ &=\sqrt{16+49}\\ &=\sqrt{65} \end{align}

If ${{z}_{1}}=1+2i$ and ${{z}_{2}}=2+3i$, evaluate $\left|\dfrac{z_1}{z_2}\right|$.

We know that $z_1=1+2i$ and $z_2=2+3i$, then \begin{align} \dfrac{z_1}{z_2}&=\dfrac{1+2i}{2+3i}\\ &=\dfrac{1+2i}{2+3i}\times \dfrac{2-3i}{2-3i}\\ &=\dfrac{\left( 2+6 \right)+\left( 4-3 \right)i}{4+9}\\ &=\dfrac{\left( 2+6 \right)+\left( 4-3 \right)i}{4+9}\\ &=\dfrac{8+i}{13} =\dfrac{8}{13}+\dfrac{1}{13}i. \end{align} Now \begin{align} \left|\dfrac{z_1}{z_2}\right|&=\sqrt{\left(\dfrac{8}{13}\right)^2+\left(\dfrac{1}{13}\right)^2}\\ &=\sqrt{\dfrac{65}{169}}\\ &=\dfrac{\sqrt{65}}{13}\end{align}